0分求调,必关

#include<iostream>
#include<cstdio>
using namespace std;
char c,t,y,z;
int n;
int main()
{
    scanf("%d",&n);
    for (int i = 0; i < n; i++)
    {
        /* code */
        y=i+'A';
        z=y+n;  
        for (char i = y; i < z; i++)
        {
            /* code */
            printf("%c",i);
        }
        printf("\n");
    }
    
	return 0;
}